398
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
(18.3)
sin nz = 2 n - 1 sinz. sin(z + rr In) ... sin[z + (n - l)rr In]
or again
(18.4)
sin rrz = 2 n - 1 II sin[(z + k)rrln].
There are analogous formulae for cos nz (replace z by z + rr 12) and, on division, for tan nz.
Suppose that n = 2m + 1 is odd and, to simplify the notation, temporarily
put
S(z) = exp rriz - exp( -rriz) = 2i sin rrz,
which will free us from having to carry around the factors 2i that would
otherwise appear. Since, in these products extended over the nth roots of
unity, the exponent k can vary mod n, one can, in (3), let it take the values
between -m and m, whence
S(nz) = i n - 1 II S(z + kin).
-m::S;k::S;m
Since a simple calculation shows that
S(x + y)S(x - y) = S(x)2 - S(y)2,
we have to group the terms k and -k for 1 :::; k :::; mj not forgetting the term
k = 0 and taking account of the fact that i n - 1 = i2m = (-1) m, we get
S(nz) = (-l)ms(z) II [S(z)2 - S(kln)2]
S(z) II [S(kln)2 - S(z)2] =
2
(
S(Z)2)
S(z) II S(kln) II 1 - S(kln)2 j
note that S(kln) = 2i. sin(krr In) does not vanish for 1 :::; k < m since
n = 2m + 1. Replacing z by zln, finally we find
(18.5)
II
2 II ( S(zl n)2)
S(z) = S(zln) S(kln)
1 - S(kln)2
.
But let us divide (5) by S(zln) and let z tend to o. Since S(z) is equivalent
to 2rriz and S(zln) to 2rrizln, the left hand side tends to n. On the right
hand side, the terms S(zln) tend to 0, so that the second product tends to 1.
Thus, in the limit,
(18.6)
II S(kln)2 = n,
which allows us to write (5) in the form